How do you sketch the curve #f(x)=1/(1+x^2)# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

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Answer 1 · 2016-12-20T07:19:12

#y in (0, 1]#. Max y = 1. Points of inflexion : #(+-1, 1/2). larr y = 0 rarr# is the asymptote.

#y = 1/(1+x^2) >= 1# and

as # x to +-oo, y to 0. S0,

#yin (1, 0]#. And so,

the global maximum is the y-intercept ( x = 0 ) 1.

At x =0,

#y'= -(2x)/(1+x^2)^2=0#, for the turning point (0, 1).

#y'' = -2/(1+x^2)^2+4x^2/(1+x^2)^3#

#=2(x^2-1)/(1+x^2)^3=0, when x = +-1. y here = 1/2.

y''' is not 0, when #x = +-1 #.

So,# (+-1, 1/2)# are the points of inflexion

graph{y(x^2+1)-1=0 [-2.5, 2.5, -1.25, 1.25]}